(a) What are the frequency and wavelength of a photon emitted during transition from n = 5 state to n = 2 state in the hydrogen atom ? (b) In which region of the electromagnetic spectrum will this radiation lie ?

(a) What are the frequency and wavelength of a photon emitted during t

1.	(a) What are the frequency and wavelength of a photon emitted during transition from n = 5 state to n = 2 state in the hydrogen atom ? (b) In which region of the electromagnetic spectrum will this radiation lie ?
Answer» Solution :(a) According to Rhdberg formula, `bar(v) = R ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2)))` Here, `R = 109, 677 cm^(-1), n_(2) = 5, n_(1) = 2` `:. bar(v) 109,677 ((1)/(2^(2)) - (1)/(5^(2))) cm^(-1) = 109,677 xx (21)/(100) cm^(-1) = 23032.2 cm^(-1)` `lamda = (1)/(v) = (1)/(23032.2 cm^(-1)) = 434 xx 10^(-7) cm= 434 xx 10^(-9) m = 434 NM` `v = (c)/(lamda) = (3 xx 10^(8) ms^(-1))/(434 xx 10^(-9)m) = 6.91 xx 10^(15) s^(-1)` (b) The wavelength, as calculated above, lies in the visible region. Otherwise too, as the jump is on the 2ND orbit, the line will belong to Balmer series and hence lie in the visible region. Alternatively, as discussed later under BOHR's MODEL, on page 2//31, energy of an electron in the nth shell is given by `E_(n) = (-21.8 xx 10^(-19))/(n^(2))J` `:.` Energy released when electron undergoes transition from n = 5 to n = 2, will be `Delta E = E_(5) - E_(2) = (-21.8 xx 10^(-19))/(5^(2)) - (- (21.8 xx 10^(19))/(2^(2))) J = -21.8 xx 10^(-19) ((1)/(25) - (1)/(4)) J` `= 21.8 xx 10^(-19) xx (21)/(100) J = 4.58 xx 10^(-19)J` But`Delta E = hv " " :. v = (Delta E)/(h)= (4.58 xx 10^(-19)J)/(6.626 xx 10^(-34)Js ) = 6.91 xx 10^(14) s^(-1) or Hz` `lamda = (c)/(v) = (3.0 xx 10^(8) ms^(-1))/(6.91 xx 10^(14) s^(-1)) = 434 xx 10^(-9) m = 434nm`